andriod09
Quadratic Equation question
How do i get from:
\[x=-11\pm \frac{121-60}{ 6 }\]
to the next part?
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andriod09
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@jwheele1
jwheele1
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ok
cos00155079
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im working on those problems too im not that good but i would simplify 121-6/ then.. which is?
jwheele1
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well first of all.....write out two different equations to eliminate that +-
andriod09
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@cos00155079 you doing the Life of Fred series?
andriod09
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you don't have to do that @jwheele1
jwheele1
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i know you dont have to but its easier...its cool we wont do that then
cos00155079
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the what? nooo...
jwheele1
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121 - 60 = ?
andriod09
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61
andriod09
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10
andriod09
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\[1\]
\[-21\]
cos00155079
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why 60/6 if it was a 61?
andriod09
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ik
jwheele1
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alright lets see now...
andriod09
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the equation is:
\[x=-11\pm \frac{ \sqrt{121-60} }{ 6 } \]
jwheele1
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oh lawd....lol
jwheele1
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sq rt of 121 = 11
jwheele1
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whats with teh line over the 0?
andriod09
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its just how it is. not my doing.
jwheele1
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well...it means that its a repeating number but its usually only after a decimal point...weird.
jwheele1
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I have no idea what to do with the 60 and the line over it.
andriod09
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the original equation is:
\[3x^{2}+11x+5=0\]
jwheele1
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any specific way she is wanting you to go about that?
andriod09
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its a messup about square roots.
jwheele1
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or can we solve for x any old way?
andriod09
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yea, using \[ax^{2}+bx+c=0\] and the
\[x=-b\pm\frac{ \sqrt{b^{2}-4ac} }{ 2a }\]
jwheele1
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the QuAdRaTiC ForMulA.....lol
jwheele1
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ok so
3x^2 + 11x +5 = 0
ax^2 + bx +c
jwheele1
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a = 3
b = 11
c = 5
jwheele1
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\[-11\pm \frac{ \sqrt{11^{2}-4(3*5c)} }{ 2(3) }\]
jwheele1
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does that look right?
jwheele1
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ignore the c next to the 5, lol
|dw:1348620564257:dw|
jwheele1
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|dw:1348620625370:dw|
jwheele1
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i see where we screwed up now in the beginning
jwheele1
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|dw:1348620687638:dw|
jwheele1
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NOW......what has your teacher said to do with this?
jwheele1
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because you can find common denominaters and put it all over one bar or you can leave it the way it is.
andriod09
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I teach my self. I am homeschooled, i use the LoF series.
jwheele1
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oh, cool
jwheele1
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well....you could work it out to this way if you wanted to....
|dw:1348620830185:dw|
andriod09
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the last answer i got was \[x_{1}\] and \[x_{2}\]
jwheele1
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OH yeah...i forgot this was all part of an original equation....this one might be too complicated for the quad formula. have you tried grouping method?
jwheele1
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i mean factoring
andriod09
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and it was \[x=\frac{-9\pm7} { 8 }\]
I did the quadratic formula on all of these so far, its a quadratic formula page. they're supposed to use the quadratic formula
jwheele1
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ok well then we must find exact values
jwheele1
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the sq rt of 61 is 7.810249675906654
jwheele1
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divided by 6 = 1.301708279317776
jwheele1
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-11 + 1.301708279317776 = ?
-11 - 1.301708279317776 = ?
andriod09
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we need not find exact values, we need to find \[x_1\] and \[x_2\]
jwheele1
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x1 and x2 are exact values
jwheele1
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it appears anyways...this one is ugly
jwheele1
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x1 = -9.698291720682224, x2 = -12.30170827931778 lol
jwheele1
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none of this seems right to me and its driving me nuts, lol
jwheele1
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for sure you need to find the square root of 61
jwheele1
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which is not what you have in your original problem at the top
jwheele1
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Heres your answer
jwheele1
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andriod09
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its these:
\[x_{1}=\frac{ -11+\sqrt{61} }{ 6 }\]
and
\[x_{1}=\frac{ -11-\sqrt{61} }{ 6 }\]
jwheele1
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thats what i would put....what you just put
andriod09
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this is what my brother says, who is in college.
jwheele1
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i am also in college. I woudl also stop there...I thought it was weird that were were trying to find the actual value for x1 and 2
andriod09
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|dw:1348621589793:dw|
jwheele1
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That is if you put it all under one common denominator which some teachers ask for...usually that is not what they want though
jwheele1
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it means the same thing but most people show it liek yoru brother did
andriod09
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w/e i have it and its done. thanks for any help!
jwheele1
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yw